## Introduction

A subsequence is a subset of the original sequence that is in the same order. For example in the string "abcdefghi", "aeg" is a subsequence but "eaq" is not because it is not in order.

The longest common subsequence between two strings A and B is the longest subsequence in A that is also in B.

For example, given A="xyaaaabcdeg", B="bcaaaaefgxy" the longest common subsequence is "aaaaeg".

xyaaaabcde_g bcaaaa___efgxy

If we try to use greedy we will see it doesn't work. For example, if we use try to take as much as B as we can, we see that we will get BCEG or if we try to take as much as A we get XY.

## Solution

Let first write a formal definition of the problem, given two strings A and B each with lengths N and M respectively, we want to find the longest common subsequence between them.

The base case is very simple: if A or B are empty strings, then trivially, the longest common subsequence is an empty string.

Now suppose that A and B are non empty strings. Let's consider the case that A and B both start with the same letter (A[0] == B[0]). It can be seen that the longest common subsequence must also begin with that letter. Thus the longest common subsequence is A[0] plus the longest common subsequence from A[1..N] and B[1..M]. However, suppose that A and B do not start with the same letter (A[0] != B[0]). Then the LCS of A and B is either the LCS of A[1..N] and B or the LCS of A and B[1..M] and we take the longer of the two. Thus, this is our recurrence relation.

### Formalization

Let lcs(A, B) be the longest common substring between A and B. Let N be the length of A and M be the length of B. Let max(A, B) be the string with longer length. lcs('', A) = '' lcs(B, '') = '' lcs(A, B) = A[0] + lcs(A[1..N], B[1..M]) if A[0] == B[0] lcs(A, B) = max(lcs(A, B[1..M]), lcs(A[1..N], B)) otherwise Example: lcs('AB', 'CAB') = max(lcs('AB','AB'), lcs('A', 'CAB')) = max('A' + lcs('B', 'B'), max(lcs('A', 'AB'), lcs('', 'CAB')) = max('A' + 'B' + lcs('', ''), max('A' + lcs('', 'B')), '')) = max('AB', max('A', '')) = max('AB', 'A') = 'AB' Example: lcs('XY', 'AB') = max(lcs('XY', 'B'), lcs('Y', 'AB')) = max(max(lcs('XY',''), lcs('Y', 'B')), max(lcs('Y','B'), lcs('', 'AB')) = ''

Note that in the second example, we are recomputing lcs('Y', 'B') twice. If we have longer strings, we will be recomputing the same values many times! We can rewrite the solution with dynamic programming by building up the solution instead of using recursion and eliminating recomputation.

Let lcs[x][y] be the least common subsequence of A[0..x] and B[0..y]. Let N be length of A and M be length of M // Base case for i from 1 to N lcs[i][0] = '' // Base case for i from 1 to M lcs[0][i] = '' for x from 1 to N for y from 1 to M if A[x - 1] == B[j - 1] lcs[x][y] = lcs[x - 1][y - 1] + A[x - 1] else lcs[x][y] = max(lcs[x - 1][y], lcs[x][y - 1] Example (string lengths): A = 'xygz' B = 'yabz' x y g z / 0 1 2 3 4 0 0 0 0 0 0 y 1 0 0 1 1 1 a 2 0 0 1 1 1 b 3 0 0 1 1 1 z 4 0 0 1 1 2

## Exercises

- Given three strings of length N, find the longest common subsequence.
- Given two strings of length N, find the longest common substring (adjacent characters).
- Given a string S, find the least number of operations (add letter, remove letter) to turn it into a palindrome.
- Example: abcdc -> abcdcb -> abcdcba (2 operations).

- Given a string A and string B, find the least number of operations (change letter, add letter, remove letter) to turn A into B.
- Example: A = abcdefh, B = bcefg, abcdefh->bcdefh->bcefh->bcefg .